Find the non real roots of the cubic equation $x^{3} - 6 x - 9 = 0$

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Solution

The correct option is B

Given equation $x^{3} - 6 x - 9$

Put x = 3 in the equation

27 - 18 - 9 = 0

(x - 3) is a one factor of the equation.

$\begin{matrix} x^{2} + 3 x + 3 x - 3 & x^{3} - 6 x - 9_x^{3}_+ 3 x^{2} 3 x^{2} - 6 x_{-} 3 x^{2}_{+} - 9 x 3 x - 9 3 x - 9 0 \end{matrix}$

$(x - 3) (x^{2} + 3 x + 3)$

Roots

$x = \frac{- 3 \pm \sqrt{9 - 12}}{2}$

$= \frac{- 3 \pm i \sqrt{3}}{2}$

Suggest Corrections

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